# Can you help me solve my mathematical problem?

1. Jun 13, 2012

### Meselwulf

I'm definitely messing up somewhere :P

The Larmor Equation is

$$\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)$$

I derived an equation stated as

$$\nabla \times \vec{F}_{ij} \cdot \hat{n} = \nabla \times \frac{\partial V(r_{ij})}{\partial r_{ij}}$$

How is put in a simpler way from the original derivation, is imagine we have what we began with

$$\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}$$

Take the dot product of the unit vector on both sides gives

$$\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}$$

Now taking the curl of F is

$$\nabla \times \vec{F}_{ij} \cdot \hat{n} = \nabla \times \frac{\partial V(r_{ij})}{\partial r_{ij}}$$

In light of this equation however

$$\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}$$

I could rewrite the Larmor energy as

$$\Delta H = \frac{2\mu}{\hbar Mc^2 e}(\vec{F}_{ij} \cdot \hat{n}) L \cdot S$$

But I am sure many agree that is not very interesting.

(and this is really starting to make my brain cells burst), taking the curl of a force gives a F/length, however, the unit vector in this equation

$$\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}$$

cancels these out and what I have again is the force again... yes? ... very circular... since this would be true, then we know what the force is anyway given earlier:

$$\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}$$

Now, just a moment ago I found the Larmor energy written as

$$\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{1}{r}\frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)$$

Notice the 1/r term which is not in my original case. If that where true, and we plug in my force example again

$$\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{1}{r}\frac{\partial V(r_{ij})}{\partial r_{ij}}\hat{n} ( L \cdot S)$$

(If I am doing this right) the unit vector would cancel out with the radius term and what would be left with is

$$\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)$$

Now if that wasn't confusing I don't know what is.

I've either had it right from the beginning, or I have made a tiny mistake which is making a huge impact on my understanding of my own equation. Any help would be gladly appreciated.

2. Jun 13, 2012

### Meselwulf

What a mess my brain got into. Assuming this is correct mind you.

I found a solution to my problem (I think). The Larmor equation is

$$\Delta H = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial V(r_{ij})}{\partial r_{ij}} ( L \cdot S)$$

What I kept deriving was:

$$\vec{F}_{ij} \cdot \hat{n} = \frac{\partial V(r_{ij})}{\partial r_{ij}}$$

What I really needed was the original derivation

$$\vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}$$

Then taking the curl of F gives

$$\nabla \times \vec{F}_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}}$$

Which removes the unit vector because nabla again is 1/length.

So what I think I have ended up with was the right derivation for the modified Larmor energy except for a dot product made on the unit vector after all, it would be

$$\Delta H = \frac{2\mu}{\hbar Mc^2 e} (\nabla \times \vec{F}_{ij}) L \cdot S$$

What a mess my brain got into. Assuming this is correct mind you.

3. Jun 13, 2012

### Meselwulf

Now, does this seem right?